From a26acf6337d2ec77595036b8e17b2d859e88e908 Mon Sep 17 00:00:00 2001
From: Sven Verdoolaege <skimo@kotnet.org>
Date: Thu, 15 Mar 2007 12:38:22 +0100
Subject: [PATCH] doc: proof of existence of suitable y for primal
 decomposition: fix notation

---
 doc/implementation.tex | 26 ++++++++++++++------------
 1 file changed, 14 insertions(+), 12 deletions(-)

diff --git a/doc/implementation.tex b/doc/implementation.tex
index b00ef45..c5ede0c 100644
--- a/doc/implementation.tex
+++ b/doc/implementation.tex
@@ -670,7 +670,9 @@ Without loss of generality, we can therefore assume for the purpose of
 showing that $R \cap S$ is non-empty that
 the $\vec u_i$ indeed form an orthogonal basis.
 
-In the orthogonal basis, the $\vec b_i^*$ are either $\vec u_i$ or $-\vec u_i$.
+In the orthogonal basis, we have $\vec b_i^* = \vec u_i$
+and the corresponding inward normal $\vec N_i$ is either
+$\vec u_i$ or $-\vec u_i$.
 Furthermore, each normal of a facet of $S$ of the first type is of the
 form $\vec {\tilde n}_{i_kj_k} = a_k \vec u_{i_k} - b_k \vec u_{j_k}$, with 
 $a_k, b_k > 0$ and ${i_k} < {j_k}$,
@@ -680,36 +682,36 @@ $a_k, b_k > 0$.
 If $\vec {\tilde n}_{i_kj_k} = a_k \vec u_{i_k} - b_k \vec u_{j_k}$
 is the normal of a facet of $S$
 then either
-$(\vec b_{i_k}^*, \vec b_{j_k}*) = (\vec u_{i_k}, \vec u_{j_k})$
+$(\vec N_{i_k}, \vec N_{j_k}) = (\vec u_{i_k}, \vec u_{j_k})$
 or
-$(\vec b_{i_k}^*, \vec b_{j_k}*) = (-\vec u_{i_k}, -\vec u_{j_k})$.
+$(\vec N_{i_k}, \vec N_{j_k}) = (-\vec u_{i_k}, -\vec u_{j_k})$.
 Otherwise, the facet would not cut $R$.
 Similarly,
 if $\vec {\tilde n}_{i_kj_k} = -a_k \vec u_{i_k} - b_k \vec u_{j_k}$
 is the normal of a facet of $S$
 then either
-$(\vec b_{i_k}^*, \vec b_{j_k}*) = (\vec u_{i_k}, -\vec u_{j_k})$
+$(\vec N_{i_k}, \vec N_{j_k}) = (\vec u_{i_k}, -\vec u_{j_k})$
 or
-$(\vec b_{i_k}^*, \vec b_{j_k}*) = (-\vec u_{i_k}, \vec u_{j_k})$.
+$(\vec N_{i_k}, \vec N_{j_k}) = (-\vec u_{i_k}, \vec u_{j_k})$.
 Assume now that $R \cap S$ is empty, then there exist
 $\lambda_k, \mu_i \ge 0$ not all zero
 such that
-$\sum_k \lambda_k \vec {\tilde n}_{i_kj_k} + \sum_l \mu_i \vec b_i^* = \vec 0$.
+$\sum_k \lambda_k \vec {\tilde n}_{i_kj_k} + \sum_l \mu_i \vec N_i = \vec 0$.
 Assume $\lambda_k > 0$ for some facet of the first type.
-If $\vec b_{j_k}^* = -\vec u_{j_k}$, then $-b_k$ can only be cancelled
+If $\vec N_{j_k} = -\vec u_{j_k}$, then $-b_k$ can only be cancelled
 by another facet $k'$ of the first type with $j_k = i_{k'}$, but then
-also $\vec b_{j_{k'}}^* = -\vec u_{j_{k'}}$.  Since the $j_k$ are strictly
+also $\vec N_{j_{k'}} = -\vec u_{j_{k'}}$.  Since the $j_k$ are strictly
 increasing, this sequence has to stop with a strictly positive coefficient
 for the largest $\vec u_{j_k}$ in this sequence.
-If, on the other hand, $\vec b_{i_k}^* = \vec u_{i_k}$, then $a_k$ can only
+If, on the other hand, $\vec N_{i_k} = \vec u_{i_k}$, then $a_k$ can only
 be cancelled by the normal of a facet $k'$ of the second kind
 with $i_k = j_{k'}$, but then
-$\vec b_{i_{k'}}^* = -\vec u_{i_{k'}}$ and we return to the first case.
+$\vec N_{i_{k'}} = -\vec u_{i_{k'}}$ and we return to the first case.
 Finally, if $\lambda_k > 0$ only for normals of facets of the second type,
-then either $\vec b_{i_k}^* = -\vec u_{i_k}$ or $\vec b_{j_k}* = -\vec u_{j_k}$
+then either $\vec N_{i_k} = -\vec u_{i_k}$ or $\vec N_{j_k} = -\vec u_{j_k}$
 and so the coefficient of one of these basis vectors will be strictly
 negative.
-That is, the sum of the normals will never by zero and
+That is, the sum of the normals will never be zero and
 the set $R \cap S$ is non-empty.
 
 For each ray $\vec u_j$ of cone $K_i$, i.e., the cone with $\vec u_i$ replaced
-- 
2.11.4.GIT